The ionization of acetic acid CH3COOH in water can be represented by the following chemical equation:CH3COOH aq + H2O l CH3COO- aq + H3O+ aq The equilibrium constant expression for this reaction is given by:Ka = [CH3COO-][H3O+] / [CH3COOH]Given that the pH of a 0.1 M acetic acid solution is 4.90, we can determine the concentration of H3O+ ions:pH = -log[H3O+]4.90 = -log[H3O+]Solving for [H3O+]:[H3O+] = 10^-4.90 1.26 x 10^-5 MSince the ionization of acetic acid produces equal amounts of CH3COO- and H3O+ ions, the concentration of CH3COO- ions is also 1.26 x 10^-5 M.Now, we can determine the concentration of CH3COOH at equilibrium. Initially, we had a 0.1 M solution of acetic acid. After ionization, the concentration of CH3COOH decreases by the amount of H3O+ ions formed:[CH3COOH] = 0.1 M - 1.26 x 10^-5 M 0.1 MNow we can plug these concentrations into the Ka expression:Ka = [ 1.26 x 10^-5 1.26 x 10^-5 ] / [0.1]Ka 1.58 x 10^-10 The equilibrium constant Ka for the ionization of acetic acid in water at 25C is approximately 1.58 x 10^-10 .